Problem: Evaluate $\int^{3\pi/2}_{3\pi/4}\sin^2\Big(\dfrac13x\Big)\,dx\,$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{6\pi+3}8$ (Choice B) B $\dfrac{3\pi+6}8$ (Choice C) C $\dfrac{6\pi-3}8$ (Choice D) D $\dfrac{3\pi-6}8$
Solution: This problem involves integrating an even power of $~\sin\,$. Since $~\cos 2A~$ can be expressed in terms of $~\sin^2A\,$, we can express $~\sin^2A~$ in terms of $~\cos2A\,$. Rewrite the integrand using the double-angle identity $ \cos2A=1-2\sin^2A $ in the form $ \sin^2A=\dfrac{1-\cos2A}2\,$. $ \int^{3\pi/2}_{3\pi/4}\sin^2\Big(\frac13x\Big)\,dx=\int^{3\pi/2}_{3\pi/4}\dfrac{1-\cos\Big(\frac23x\Big)}2\,dx$ Now we can do the integration. $ \begin{aligned}\int^{3\pi/2}_{3\pi/4}\dfrac{1-\cos\Big(\frac23x\Big)}2\,dx&=\frac12\int_{3\pi/4}^{3\pi/2}\Bigg(1-\cos\Big(\frac23x\Big)\Bigg)dx\\ \\ \\&=\dfrac12\Bigg(x-\dfrac32\sin\Big(\dfrac23x\Big)\Bigg)\Bigg|^{3\pi/2}_{3\pi/4}\\ \\ \\&=\frac12\Bigg(\Big(\frac{3\pi}2-\frac32\sin\pi\Big)-\Big(\frac{3\pi}4-\frac32\sin\frac\pi2\Big)\Bigg)\\ \\ \\&=\frac12\Big(\frac{3\pi}2-0-\frac{3\pi}4+\frac32\Big)\\ \\ \\&=\frac12\Big(\frac{3\pi}4+\frac32\Big)\\ \\ \\&=\dfrac{3\pi+6}8\end{aligned}$